Say that I have any homomorphisms of commutative rings, $A \rightarrow B, A \rightarrow C.$ I recently read that $B \otimes_A C$ is the pushout of the morphisms in the category of commutative rings. However, I understood tensor products as defined for modules over a commutative ring. Can we somehow realize $B, C$ as modules over $A$ via the homomorphisms or is $B \otimes_A C$ as defined by the pushout a generalization of the 'normal' definition of tensors? It seems unlikely that it is a generalization as it seems to depend on these morphisms whereas the tensors of modules doesn't depend on any sort of morphisms.
Tensor product terminology in category theory
category-theorytensor-productsterminology
Best Answer
Take all rings here to be commutative. A ring homomorphism $f:A\to B$ makes $B$ into an $A$-module. In detail, the module action is $a\cdot b=f(a)b$. With another ring homomorphism $g:A\to C$ then we have two $A$-modules, and can form the tensor product $B\otimes_A C$.
At first $B\otimes_A C$ is just a module. But it has a multiplication, defined as the composition $$(B\otimes_A C)\times(B\otimes_A C)\to (B\otimes_A C)\otimes_A (B\otimes_A C) \to (B\otimes_A B)\otimes_A (C\otimes_A C)\to B\otimes_A C.$$ The middle map is just permuting the factors, and the last map is induced by $(b\otimes b')\otimes(c\otimes c')\mapsto bb'\otimes cc'$. In terms of elements: $$(b\otimes c)(b'\otimes c')=bb'\otimes cc'.$$
Then $B\otimes_A C$ is a ring. There are ring homomorphisms from $B$ and $C$ to it, the first given by $a\mapsto a\otimes 1_C$. Now one sits down with a large sheet of paper, and proves that the map $A\to B\otimes_A C$ is the pushout of $A\to B$ and $A\to C$.